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Divisible By 11

Problem

Consider the following results.

101 + 1 = 11
102 minus 1 = 99 = 9 times 11
103 + 1 = 1001 = 91 times 11
104 minus 1 = 9999 = 909 times 11
105 + 1 = 100001 = 9091 times 11

Prove that 10nminus1 is divisible by 11 if n is even and 10n+1 is divisible by 11 if n is odd.


Solution

As (xminus1)n = (xminus1)(xnminus1 + xxminus2 + ... + 1), we get:

102kminus1 = 100kminus1 = (100minus1)(100kminus1 + ... + 1) = 99Q congruent 0 mod 11.

That is, 10nminus1 is divisible by 11 if n is even.

As 102k congruent 1 mod 11, we can multiply both sides of the congruence by 10 to obtain 102k+1 congruent 10 congruent -1 mod 11; that is, 102k+1+1 congruent 0 mod 11.

Hence we prove that 10n+1 is divisible by 11 if n is odd.

Prove that 100n+1 is divisible by 101 iff n is odd, 1000n+1 is divisible by 1001 iff n is odd.
What about 10000n+1? Generalise.

Problem ID: 208 (17 Feb 2005)     Difficulty: 3 Star

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